The correct answer is option D which is p ³+ 2; when p = 3 the number of plants is 29.
The complete question is given below:-
liana cubed the number of flowering plants in her garden, then added 2 vegetable plants. Let p represent the original number of plants in her garden. Which shows an expression to represent the total number of plants in her garden and the total number of plants if p = 3? 3 p minus 2; when p = 3 the number of plants is 7 3 p + 2; when p = 3 the number of plants is 11 p cubed minus 2; when p = 3 the number of plants is 25 p cubed + 2; when p = 3 the number of plants is 29
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What is an expression?</h3>
Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.
The given expression is liana cubed the number of flowering plants in her garden, then added 2 vegetable plants.
If she cubed the number of the plants will be:- P³
Now she added two vegetable plants:- 2
Then the equation will be:- p³ + 2. Now we will put the value of p = 3 in the equation.
Total number of the plants = ( 3 )³ + 2 = 27 + 2 = 29
Therefore the correct answer is option D which is p³+ 2; when p = 3 the number of plants is 29.
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Answer:
Step-by-step explanation:
the 3rd one
Answer:
Her error was multiplying -3 and -4 which gives +12 but she wrote -12 instead
Step-by-step explanation:
Mariana workings:
-3(2x - 4) = -12
-6x - 12 = -12
Her error was multiplying -3 and -4 which gives +12 but she wrote -12 instead
-6x = 0
x=0
Correct workings:
-3(2x - 4) = -12
-6x + 12 = - 12
-6x = -12 - 12
-6x = -24
x = -24/-6
x = 4
Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴

⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =

∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)